Experimental Design B/C

[Anomaly]

Click to Show Answer

1. at least 3

2. at least 4

Actually, only 3 CVs are needed.

Oh, then they changed it, it was 4 last year.

Yes, if you look at the 2019 season rubric it'll say you only need 3 CVs now

[Jacobi]

dxu46 can go.

[dxu46]

You test the affect of running speed on perspiration. You results are, for 4 mph, 67 mL, 69 mL, and 81 mL. For 6 mph, your results are 87, 82, and 91 mL. For 8 mph, your results are 101, 101, and 109 mL. List and calculate at least 6 useful statistics.

(disclaimer: this is a FICTIONAL experiment. The results are highly improbable, and the experiment is unlikely to be done. But the question is about statistics, not ideas.)

[Jacobi]

You test the affect of running speed on perspiration. You results are, for 4 mph, 67 mL, 69 mL, and 81 mL. For 6 mph, your results are 87, 82, and 91 mL. For 8 mph, your results are 101, 101, and 109 mL. List and calculate at least 6 useful statistics.

(disclaimer: this is a FICTIONAL experiment. The results are highly improbable, and the experiment is unlikely to be done. But the question is about statistics, not ideas.)

Do they have to be all different?

6 Different Stats (Click to Show Hidden Text)

1. Mean Perspiration (mL): 4 mph -> 72 mL (= (67 + 69 + 81)/3), 6 mph -> 87 mL, 8 mph -> 104 mL
2. Standard Deviation Perspiration (mL): 4 mph -> [math]\sqrt{\frac{(67-72)^2 + (69 - 72)^2 + (81 - 72)^2)}{3}} =[/math] 7.6 mL, 6 mph -> 4.5 mL, 8 mph -> 4.6 mL
3. Median Perspiration (mL): 4 mph -> [b]Find middle of 3 values in order, that is the second-highest value:[/b] [math]67, \underline{69}, 81[/math] 69 mL, 6 mph -> 87 mL, 8 mph - > 101 mL
4. Range Perspiration (mL): 4 mph -> [math]R = max - min = 81-67 =[/math] 14 mL, 6 mph -> 9 mL, 8mph -> 8 mL.
5. Mean Absolute Deviation (mL): 4 mph -> [math]MAD = \frac{1}{n}\Sigma_i |x_i - \overline{x}| = (|67 - 72.33| + |69 - 72.33| + |81 - 72.33|) / 3 = 5.8 mL[/math], 6 mph -> 3.1 mL, 8 mph -> 3.6 mL
6. Pearson Correlation between Hours and Milliliters: [math]r = \Sigma_i \frac{(x_i - \overline{x})(y_i - \overline{y})}{s_x s_y}[/math]
[math]\overline{x} = 6.00[/math]
[math]\overline{y} = 87.6[/math]
[math]s_y = 40.9[/math]
[math]s_x = 4.9[/math]
.
.
.
[math]r = 0.938[/math]

[dxu46]

You test the affect of running speed on perspiration. You results are, for 4 mph, 67 mL, 69 mL, and 81 mL. For 6 mph, your results are 87, 82, and 91 mL. For 8 mph, your results are 101, 101, and 109 mL. List and calculate at least 6 useful statistics.

(disclaimer: this is a FICTIONAL experiment. The results are highly improbable, and the experiment is unlikely to be done. But the question is about statistics, not ideas.)

Do they have to be all different?

6 Different Stats (Click to Show Hidden Text)

1. Mean Perspiration (mL): 4 mph -> 72 mL (= (67 + 69 + 81)/3), 6 mph -> 87 mL, 8 mph -> 104 mL
2. Standard Deviation Perspiration (mL): 4 mph -> [math]\sqrt{\frac{(67-72)^2 + (69 - 72)^2 + (81 - 72)^2)}{3}} =[/math] 7.6 mL, 6 mph -> 4.5 mL, 8 mph -> 4.6 mL
3. Median Perspiration (mL): 4 mph -> [b]Find middle of 3 values in order, that is the second-highest value:[/b] [math]67, \underline{69}, 81[/math] 69 mL, 6 mph -> 87 mL, 8 mph - > 101 mL
4. Range Perspiration (mL): 4 mph -> [math]R = max - min = 81-67 =[/math] 14 mL, 6 mph -> 9 mL, 8mph -> 8 mL.
5. Mean Absolute Deviation (mL): 4 mph -> [math]MAD = \frac{1}{n}\Sigma_i |x_i - \overline{x}| = (|67 - 72.33| + |69 - 72.33| + |81 - 72.33|) / 3 = 5.8 mL[/math], 6 mph -> 3.1 mL, 8 mph -> 3.6 mL
6. Pearson Correlation between Hours and Milliliters: [math]r = \Sigma_i \frac{(x_i - \overline{x})(y_i - \overline{y}}{s_x s_y}[/math]
[math]\overline{x} = 6.00[/math]
[math]\overline{y} = 87.6[/math]
[math]s_y = 40.9[/math]
[math]s_x = 4.9[/math]
.
.
.
[math]r = 0.938[/math]

Your turn.

[Jacobi]

Standard deviation measures what quality of a data distribution?

[OrigamiPlanet]

Standard deviation measures what quality of a data distribution?

Answer (Click to Show Hidden Text)

It measures the average variation that the actual values have from the mean value.

[Jacobi]

Standard deviation measures what quality of a data distribution?

Answer (Click to Show Hidden Text)

It measures the average variation that the actual values have from the mean value.

Not exactly...
Try again.

[UTF-8 U+6211 U+662F]

Standard deviation measures what quality of a data distribution?

Answer (Click to Show Hidden Text)

It measures the average variation that the actual values have from the mean value.

Not exactly...
Try again.

Wait, what's wrong with that answer? That's pretty much the intent of a standard deviation. Did you just want the spread of the distribution?

Edit: Also, correcting this because the mismatched parentheses bother me...

You test the affect of running speed on perspiration. You results are, for 4 mph, 67 mL, 69 mL, and 81 mL. For 6 mph, your results are 87, 82, and 91 mL. For 8 mph, your results are 101, 101, and 109 mL. List and calculate at least 6 useful statistics.

(disclaimer: this is a FICTIONAL experiment. The results are highly improbable, and the experiment is unlikely to be done. But the question is about statistics, not ideas.)

Do they have to be all different?

6 Different Stats (Click to Show Hidden Text)

1. Mean Perspiration (mL): 4 mph -> 72 mL (= (67 + 69 + 81)/3), 6 mph -> 87 mL, 8 mph -> 104 mL
2. Standard Deviation Perspiration (mL): 4 mph -> [math]\sqrt{\frac{(67-72)^2 + (69 - 72)^2 + (81 - 72)^2}{3}} =[/math] 7.6 mL, 6 mph -> 4.5 mL, 8 mph -> 4.6 mL
3. Median Perspiration (mL): 4 mph -> [b]Find middle of 3 values in order, that is, the second-highest value:[/b] [math]67, \underline{69}, 81[/math] 69 mL, 6 mph -> 87 mL, 8 mph - > 101 mL
4. Range Perspiration (mL): 4 mph -> [math]R = max - min = 81-67 =[/math] 14 mL, 6 mph -> 9 mL, 8mph -> 8 mL.
5. Mean Absolute Deviation (mL): 4 mph -> [math]\textrm{MAD} = \frac{1}{n}\Sigma_i |x_i - \overline{x}| = (|67 - 72.33| + |69 - 72.33| + |81 - 72.33|) / 3 = 5.8 mL[/math], 6 mph -> 3.1 mL, 8 mph -> 3.6 mL
6. Pearson Correlation between Hours and Milliliters: [math]r = \Sigma_i \frac{(x_i - \overline{x})(y_i - \overline{y}}{s_x s_y}[/math]
[math]\overline{x} = 6.00[/math]
[math]\overline{y} = 87.6[/math]
[math]s_y = 40.9[/math]
[math]s_x = 4.9[/math]
.
.
.
[math]r = 0.938[/math]

should be

6 Different Stats (Click to Show Hidden Text)

1. Mean Perspiration (mL): 4 mph -> [math]\frac{67+69+81}{3} =[/math] 72 mL, 6 mph -> 87 mL, 8 mph -> 104 mL
2. Standard Deviation Perspiration (mL): 4 mph -> [math]\sqrt{\frac{(67-72)^2 + (69 - 72)^2 + (81 - 72)^2)}{3}} =[/math] 7.6 mL, 6 mph -> 4.5 mL, 8 mph -> 4.6 mL
3. Median Perspiration (mL): 4 mph -> [b]Find middle of 3 values in order, that is the second-highest value:[/b] [math]67, \underline{69}, 81 \Rightarrow[/math] 69 mL, 6 mph -> 87 mL, 8 mph - > 101 mL
4. Range Perspiration (mL): 4 mph -> [math]R = \textrm{max} - \textrm{min} = 81-67 =[/math] 14 mL, 6 mph -> 9 mL, 8mph -> 8 mL.
5. Mean Absolute Deviation (mL): 4 mph -> [math]MAD = \frac{1}{n}\Sigma_i |x_i - \overline{x}| = \frac{|67 - 72.33| + |69 - 72.33| + |81 - 72.33|}{3} =[/math] 5.8 mL, 6 mph -> 3.1 mL, 8 mph -> 3.6 mL
6. Pearson Correlation between Hours and Milliliters: [math]r = \Sigma_i \frac{(x_i - \overline{x})(y_i - \overline{y})}{s_x s_y}[/math]
[math]\overline{x} = 6.00[/math]
[math]\overline{y} = 87.6[/math]
[math]s_y = 40.9[/math]
[math]s_x = 4.9[/math]
.
.
.
[math]r = 0.938[/math]

[Jacobi]

Answer (Click to Show Hidden Text)

It measures the average variation that the actual values have from the mean value.

Not exactly...
Try again.

Wait, what's wrong with that answer? That's pretty much the intent of a standard deviation. Did you just want the spread of the distribution?

Edit: Also, correcting this because the mismatched parentheses bother me...

You test the affect of running speed on perspiration. You results are, for 4 mph, 67 mL, 69 mL, and 81 mL. For 6 mph, your results are 87, 82, and 91 mL. For 8 mph, your results are 101, 101, and 109 mL. List and calculate at least 6 useful statistics.

(disclaimer: this is a FICTIONAL experiment. The results are highly improbable, and the experiment is unlikely to be done. But the question is about statistics, not ideas.)

Do they have to be all different?

6 Different Stats (Click to Show Hidden Text)

1. Mean Perspiration (mL): 4 mph -> 72 mL (= (67 + 69 + 81)/3), 6 mph -> 87 mL, 8 mph -> 104 mL
2. Standard Deviation Perspiration (mL): 4 mph -> [math]\sqrt{\frac{(67-72)^2 + (69 - 72)^2 + (81 - 72)^2}{3}} =[/math] 7.6 mL, 6 mph -> 4.5 mL, 8 mph -> 4.6 mL
3. Median Perspiration (mL): 4 mph -> [b]Find middle of 3 values in order, that is, the second-highest value:[/b] [math]67, \underline{69}, 81[/math] 69 mL, 6 mph -> 87 mL, 8 mph - > 101 mL
4. Range Perspiration (mL): 4 mph -> [math]R = max - min = 81-67 =[/math] 14 mL, 6 mph -> 9 mL, 8mph -> 8 mL.
5. Mean Absolute Deviation (mL): 4 mph -> [math]\textrm{MAD} = \frac{1}{n}\Sigma_i |x_i - \overline{x}| = (|67 - 72.33| + |69 - 72.33| + |81 - 72.33|) / 3 = 5.8 mL[/math], 6 mph -> 3.1 mL, 8 mph -> 3.6 mL
6. Pearson Correlation between Hours and Milliliters: [math]r = \Sigma_i \frac{(x_i - \overline{x})(y_i - \overline{y}}{s_x s_y}[/math]
[math]\overline{x} = 6.00[/math]
[math]\overline{y} = 87.6[/math]
[math]s_y = 40.9[/math]
[math]s_x = 4.9[/math]
.
.
.
[math]r = 0.938[/math]

should be

6 Different Stats (Click to Show Hidden Text)

1. Mean Perspiration (mL): 4 mph -> [math]\frac{67+69+81}{3} =[/math] 72 mL, 6 mph -> 87 mL, 8 mph -> 104 mL
2. Standard Deviation Perspiration (mL): 4 mph -> [math]\sqrt{\frac{(67-72)^2 + (69 - 72)^2 + (81 - 72)^2)}{3}} =[/math] 7.6 mL, 6 mph -> 4.5 mL, 8 mph -> 4.6 mL
3. Median Perspiration (mL): 4 mph -> [b]Find middle of 3 values in order, that is the second-highest value:[/b] [math]67, \underline{69}, 81 \Rightarrow[/math] 69 mL, 6 mph -> 87 mL, 8 mph - > 101 mL
4. Range Perspiration (mL): 4 mph -> [math]R = \textrm{max} - \textrm{min} = 81-67 =[/math] 14 mL, 6 mph -> 9 mL, 8mph -> 8 mL.
5. Mean Absolute Deviation (mL): 4 mph -> [math]MAD = \frac{1}{n}\Sigma_i |x_i - \overline{x}| = \frac{|67 - 72.33| + |69 - 72.33| + |81 - 72.33|}{3} =[/math] 5.8 mL, 6 mph -> 3.1 mL, 8 mph -> 3.6 mL
6. Pearson Correlation between Hours and Milliliters: [math]r = \Sigma_i \frac{(x_i - \overline{x})(y_i - \overline{y})}{s_x s_y}[/math]
[math]\overline{x} = 6.00[/math]
[math]\overline{y} = 87.6[/math]
[math]s_y = 40.9[/math]
[math]s_x = 4.9[/math]
.
.
.
[math]r = 0.938[/math]

What's wrong is the standard deviation measures the typical, not average, deviation from the mean. It's a fine but critical distinction. What parentheses did you correct?